Solve for $x$ and $y$ by deriving an expression for $y$ from the second equation, and substituting it back into the first equation. $\begin{align*}-3x-5y &= 4 \\ 5x+3y &= -8\end{align*}$
Solution: Begin by moving the $x$ -term in the second equation to the right side of the equation. $3y = -5x-8$ Divide both sides by $3$ to isolate $y$ $y = {-\dfrac{5}{3}x - \dfrac{8}{3}}$ Substitute this expression for $y$ in the first equation. $-3x-5({-\dfrac{5}{3}x - \dfrac{8}{3}}) = 4$ $-3x + \dfrac{25}{3}x + \dfrac{40}{3} = 4$ Simplify by combining terms, then solve for $x$ $\dfrac{16}{3}x + \dfrac{40}{3} = 4$ $\dfrac{16}{3}x = -\dfrac{28}{3}$ $x = -\dfrac{7}{4}$ Substitute $-\dfrac{7}{4}$ for $x$ back into the top equation. $-3( -\dfrac{7}{4})-5y = 4$ $\dfrac{21}{4}-5y = 4$ $-5y = -\dfrac{5}{4}$ $y = \dfrac{1}{4}$ The solution is $\enspace x = -\dfrac{7}{4}, \enspace y = \dfrac{1}{4}$.